For a thin lens, prove that 1 f = μ - 1 1 R 1 - 1 R 2

The given equation is of the Lens Maker's Formula, to prove this, let us take any lens made up of two surfaces A B C and A D C of radii of curvatures R 1 and R 2 . A point object is placed on the principal axis at O . Image of an object O is formed at I due to A B and this image acts as an object for the second surface A D and its image produced at I 1 . Let the distance of the object O from the lens is u and the image I 1 is at distance v 1 , and Image I is at distance v . From the first surface A B C image of O is obtained at I , that way by the formula for refraction at spherical surface, μ - 1 R 1 = μ v 1 - 1 u . . . . . . . . . . . . i Here, μ is the refractive index which is the ratio of the speed of light in a vacuum to its speed in a specific medium. From the second surface A D C image of I is obtained at I 1 ,by the formula for refraction at spherical surface, 1 μ - 1 R 2 = 1 μ - 1 v - 1 v 1 . . . . . . . . . . i i Here, μ = 1 u , u = v 1 for A D surface because light ray passing from denser to rarer medium Now, on multiplying μ with equation i i 1 v - μ v 1 = - 1 - μ R 2 . . . . . . . . . . . . i i i Now, adding equation i & i i i 1 v - 1 u = μ - 1 R 1 + 1 - μ R 2 . . . . . . . . . . i v 1 v - 1 u = μ - 1 R 1 - μ - 1 R 2 We know that when u = ∞ then v = f 1 f - 1 ∞ = μ - 1 1 R 1 - 1 R 2 Now, we know that 1 ∞ = 0 So, 1 f = μ - 1 1 R 1 - 1 R 2 Hence, Proved.

For a thin lens, prove that 1 f = μ - 1 1 R 1 - 1 R 2

The given equation is of the Lens Maker's Formula, to prove this, let us take any lens made up of two surfaces A B C and A D C of radii of curvatures R 1 and R 2 . A point object is placed on the principal axis at O . Image of an object O is formed at I due to A B and this image acts as an object for the second surface A D and its image produced at I 1 . Let the distance of the object O from the lens is u and the image I 1 is at distance v 1 , and Image I is at distance v . From the first surface A B C image of O is obtained at I , that way by the formula for refraction at spherical surface, μ - 1 R 1 = μ v 1 - 1 u . . . . . . . . . . . . i Here, μ is the refractive index which is the ratio of the speed of light in a vacuum to its speed in a specific medium. From the second surface A D C image of I is obtained at I 1 ,by the formula for refraction at spherical surface, 1 μ - 1 R 2 = 1 μ - 1 v - 1 v 1 . . . . . . . . . . i i Here, μ = 1 u , u = v 1 for A D surface because light ray passing from denser to rarer medium Now, on multiplying μ with equation i i 1 v - μ v 1 = - 1 - μ R 2 . . . . . . . . . . . . i i i Now, adding equation i & i i i 1 v - 1 u = μ - 1 R 1 + 1 - μ R 2 . . . . . . . . . . i v 1 v - 1 u = μ - 1 R 1 - μ - 1 R 2 We know that when u = ∞ then v = f 1 f - 1 ∞ = μ - 1 1 R 1 - 1 R 2 Now, we know that 1 ∞ = 0 So, 1 f = μ - 1 1 R 1 - 1 R 2 Hence, Proved.

A candle is placed at 12 cm distance from a concave mirror. Radius of curvature of the mirror is 10 cm . Find the position, nature and magnification of the image.

It is given that: Distance of candle from the mirror, u = - 12 cm The radius of curvature of the mirror, R = - 10 cm The focal length of the mirror, f = R / 2 ⇒ f = - 10 / 2 cm = - 5 cm By mirror formula, 1 f = 1 v + 1 u 1 - 5 = 1 v + 1 - 12 1 v = 1 - 5 + 1 12 1 v = - 12 - 5 60 = - 7 60 ⇒ v = - 8 . 57 cm Thus, the image will be formed at 8 . 57 cm in front of the mirror. The image formed will be real and inverted. The magnification of mirror, m = - v u m = - 60 / 7 - 10 = 6 7 = 0 . 86 The image formed is small compared to the object.

Describe the terrestrial telescope under the following points: Ray diagram of the formation of image.

A telescope is an optical instrument that helps to see far away objects clearly. A telescope is used to see celestial bodies like stars, moons and other planets. A terrestrial telescope is a refracting telescope that works on the principle of refraction. For this, it uses a lens. A terrestrial telescope uses three convex (converging) lenses as you can see in the ray diagram below. The three lenses are called the objective lens, the erecting lens and the eye lens. The rays of light from a very far object are incident on the objective lens. They refract through the objective lens and intersect at a point on the other side of the lens. Due to this an intermediate image (A’B’) of the object is formed between the erecting lens and the objective lens. The image A’B; acts as an object for the erecting lens and one more intermediate image (A’’B’’) is formed on the other side of this lens. However, this image formed is inverted as shown. Lastly, the rays of light refract through the eye lens and a final image of the object is formed. Final image formed is erect and enlarged and this can be seen through the eye lens. In this way, a terrestrial telescope formed an enlarged image of a very far away object.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2019>Q 9

HARD

12th Chhattisgarh Board

IMPORTANT

Earn 100

Draw a labelled diagram depicting the formation of image by an Astronomical Telescope in normal adjustment (when final image is formed at infinity

Important Questions on Ray Optics and Optical Instruments

EASY

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2018>Q 2

The ratio of angular colour dispersion to the angle of deviation for the mean colour is known as _____.

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2019>Q 10

With the help of a ray diagram, show the formation of the image of a point object by the refraction of light at a spherical convex surface separating two media of refraction indices

μ_{1}

and

μ_{2}

(

μ_{2} > μ_{1}

) respectively, Using this diagram derive the relation:

\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}

Where used symbols have their usual meanings

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2019>Q 11

Draw the ray diagram showing the refraction of monochromatic light through a glass prism of refraction angle

A

. Hence, derive the formula for the refractive index

μ

of glass prism in terms of angle of prism and angle of minimum deviation. Also, draw

i - δ

curve of prism.

EASY

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2020>Q 12

Which mirror has an infinite focal length?

EASY

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2020>Q 13

What is optical fibre? Write its two important uses.

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2020>Q 14

For a thin lens, prove that

\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2020>Q 15

Draw the ray diagram to show the formation of the image by compound microscope and obtain an expression for the magnifying power when the final image is formed at least distance of distinct vision.

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